Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
KingGeorge
Group Title
[SOLVED] Here's a cute predicate logic problem.
Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).
 2 years ago
 2 years ago
KingGeorge Group Title
[SOLVED] Here's a cute predicate logic problem. Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).
 2 years ago
 2 years ago

This Question is Closed

experimentX Group TitleBest ResponseYou've already chosen the best response.0
is xyzw a combination of digit or multiplication of digit??
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
multiplication
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
lol .. didn't realize they were complex
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Just prove/disprove it for \(\mathbb{R}\) and it will become clear why it doesn't matter if they're complex numbers.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Hint below in white LaTeX\[\color{white}{\text{There is a certain value for y that you should choose to prove it.}}\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\[ \color{white}{\text{LOL i didn't know that!!.}} \]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Another, probably more helpful hint, below in white LaTeX\[\color{white}{\text{What value of y can you choose so that xyzw is constant?}}\]
 2 years ago

nbouscal Group TitleBest ResponseYou've already chosen the best response.2
Choose y=0, then choose w=(x+z).
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
\(\exists\, y\) means that we can choose a specific value for y. It's a variable, but one that we have complete control over.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
then it must be false ... i guess so!!
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
The "there exists" only means that at least one possible value of y where this is true. One value that we know works is \(y=0\). This means that there does exists at least one value for \(y\) that makes the statement true, so we have proven that the original statement is true.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
does't y=0 make statement untrue?? xyzw=0, for y=0
 2 years ago

nbouscal Group TitleBest ResponseYou've already chosen the best response.2
Right, both the sum and the multiplication end up equaling zero because you set w=(x+z).
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
If \(y\) is 0, and \(x=(x+z)\) we have \[x+0+z+((x+z))=x+zxz=0\]and \[x\cdot0\cdot z\cdot w=0\]So that means that \[x+y+z+w=xyzw\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
Oh ... stupid me ... i never considered the negative values!!
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Hope everyone enjoyed this little (unintuitive) problem. Good night.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
thanks!! i'll try to think broader way next time!!
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
I'll try to get a problem just for you next time then. Any preference about the subject?
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.